Identify the coefficient of y or the number in front of y which is '8' in this case then look for two numbers whose sum is 8 and whose product is 12

6 x 2 = 12 6 +2 = 8

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[3]

Since 6 and 2 are the two numbers we were looking for in the previous step
and since y is the variable of the equation ,we multiply both 6 and 2 by y ,i.e
6 x y = 6y 2 x y = 2y
now since 6y + 2y = 8y we replace 8y with 6y + 2y

8y = 6y + 2y

4y2 + 6y + 2y + 3 = 0

[4]

Now put the terms into groups of two

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( 4y2 + 6y ) + ( 2y + 3 )= 0

[5]

Factorise each group ,choosing the HCF of each group ,putting the HCF
of each group outside the brackets dividing both terms in the group by the HCF of that group

HCF of 4y2 and 6y is 2y and the HCF of 2y and 3 is 1

2y[(4y2/2y) + (6y/2y)] + 1[(2y/1) + (3/1)]= 0

[6]

Simplify left hand side of equation obtained in previous step

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2y(2y + 3) + 1(2y + 3)= 0

[7]

now since 2y+3 is a common factor of the terms on left hand side of equation obtained in step 6 ,factor out 2y+3

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(2y+3)[(2y(2y + 3)/(2y+3))+ (1(2y + 3)/(2y+3))]= 0

[8]

Simplify left hand side of equation obtained in previous step
Note: in the previous step you donot have to show how you factorise 2y+3 ,
you can just take (2y+3) since it appears in both terms in step [6] and multiply it to the sum of the numbers outside each brackets (paying attention to their sign(+/-) ).

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(2y+3)(2y+1)= 0

[9]

Now set each factor or group in equation obtained in step[8] equal to zero